AIPMT 2008 Question on Magnetic force

Most of you might have noted that a magnetic dipole placed in a non uniform magnetic field generally experiences a net force and a torque. If the magnetic field is uniform, the net force will be zero; but there will be a torque (if the magnetic moment vector is not parallel or antiparallel to the magnetic field direction). The following question which appeared in AIPMT 2008 question paper is worth noting:

A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are F₁, F₂ and F₃ respectively and are in the same plane of the paper and along the directions shown, the force on the segment QP is

(1) √[(F₃ – F₁)² + F₂²]

(2) √[(F₃ – F₁)² – F₂²]

(3) F₃ – F₁ + F₂

(4) F₃ – F₁ – F₂

The net force on the loop in the horizontal direction is F₃ – F₁. Since the force F₂ is in the vertical direction, the reultant of the three forces F₁, F₂ and F₃ is √[(F₃ – F₁)² + F₂²]. Since the net force on the entire loop must be zero in the uniform magnetic field, the force on the segment QP must be the equilibrant of √[(F₃ – F₁)² + F₂²]. Therefore the force on the segment QP has magnitude √[(F₃ – F₁)² + F₂²] and direction opposite [Option (1)].

Now consider a question which is a little more difficult:

A proton of charge q and mass m proceeding along the positive X-direction with speed v encounters a uniform magnetic field of flux density B directed along the negative Z-direction. If the field is confined in the region between x = 0 and x = d and the proton emerges from the field along a direction making an angle of 45º with its initial velocity, the value of d must be

(a) (1/√2)(mv/qB)

(b) √2 mv/qB

(c) mv/qB

(d) 2 mv/qB

(e) mv/2qB

The situation is shown in the adjoining figure. C is the centre of the circular path of the proton in the magnetic field and AB is the radius R drawn from the point A from which the proton exits from the field. We have

R = mv/qB which you get by equating the centripetal force mv²/R to the magnetic force qvB.

Since d = AN = R sin45º where N is the foot of the perpendicular (drawn from A) to the Y-axis, we have

D = R/√2 = (1/√2)(mv/qB)

You will find many useful questions (with solution) on magnetic force at apphysicsresources

A Question (MCQ) Involving Lorentz Force

Here is an interesting question which requires some imagination on your part:

Parallel Electric and magnetic fields act at a certain of space. If a positive point charge is projected into this region with a velocity ‘v’, making a small angle with respect to the fields, the path of the point charge within the fields will be

(a) a straight line

(b) a helix of constant pitch and radius

(c) a helix of increasing pitch and radius

(d) a helix of constant pitch and increasing radius

(e) a helix of increasing pitch and constant radius

Since the charge is projected at a small angle, it will move along a helical path. This is because of the fact that the velocity component perpendicular to the magnetic field forces the particle to move along a circle and the velocity component parallel to the magnetic field makes it move forward in the direction of the magnetic field. In the absence of the electric field, the particle will move along a helix of constant pitch and radius.

Since there is an electric field parallel to the magnetic field, the forward component of the velocity goes on increasing while the perpendicular component remains unchanged in magnitude. The particle will therefore move along a helix of increasing pitch and constant radius under the combined action of the electric and magnetic Lorentz forces.

Questions involving Magnetic Fields

The following M.C.Q. which appeared in the All India Pre –Medical/Dental Entrance-2005 (C.B.S.E) test paper will be interesting to you:
An electron moves in a circular orbit with a uniform speed ‘v’. It produces a magnetic field ‘B’ at the centre of the circle. The radius of the circle is proportional to
(a) √(B/v) (b) B/v (c) √(v/B) (d) v/B
The electron, moving along the circular path is equivalent to a current I = e/T = ev/(2πr) where ‘e’ is the electronic charge, T is the period, ‘v’ is the speed and ‘r’ is the radius of the circular motion. The magnetic field at the centre of the circle, B = μ₀I/2r = μ₀ev/4πr² so that r α √(v/B). The correct option is (c).
Now, consider the following simple question. (Be careful however, so that you won’t pick out a wrong answer):
A long straight conductor P carries a current ‘I’ while another parallel long straight conductor Q carries a current 2I in the same direction. The magnetic field midway between them is B. If the current 2I in the conductor Q is switched off, the magnetic field midway between the conductors will be
(a) B/3 (b) B (c) –B (d) –B/2 (e) 2B
The magnetic fields produced by the conductors are in opposition in the region between them. The field produced by Q is twice the field produced by P. If the field produced by Q is 2B, the field produced by P is –B. This is indeed the case, since the net field is B when both currents are switched on. When the current 2I in Q is switched off, the field due to Q alone is present. The answer therefore is –B [Option (c)].
The following question also is simple. It is designed to check your understanding of certain basic things:
A straight conductor of length ‘L’ carrying a current ‘I’ is bent in the form of a semicircle. The magnetic flux density (in tesla) at the centre of the semicircle is approximately
(a) 10^-6I/L (b) 10-7I/L (c) 10^-6π²I/L (d) 10⁷π²I/L (e) π²I/L
The magnetic field at the centre of a single turn circular current carrying coil is μ₀I/2r and hence the magnetic field at the centre of a semicircle is μ₀I/4r. The radius of the semicircle obtained by bending the straight conductor of length L is, r = L/π. On substituting this value, the magnetic field at the centre of the semicircle is μ₀πI/ 4L. Since μ₀ = 4π ×10^-7, the field is 10^-7π² I/L. But π² is approximately equal to 10 so that the answer is 10^-6I/L [Option (a)].
Let us now consider the following M.C.Q. which appeared in the Karnataka Common Entrance test paper of 2005:
The electrons in the beam of a television tube move horizontally from south to north. The vertical component of the earth’s magnetic field points down. The electron is deflected towards
(a) west (b) no deflection (c) east (d) north to south.
It will be convenient to use Fleming’s left hand rule here. You have to hold the fore-finger, middle finger and thumb of your left hand along mutually perpendicular directions so that the fore-finger points along the magnetic field and the middle finger along the direction of the conventional current (and hence opposite to the direction of the electron). The thumb will then give you the direction of deflection, which you can easily obtain as along the east [Option (c)]. You will find more multiple choice questions (with solution) at physicsplus: Magnetic Force on Moving Charges